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MTLE-6120: Advanced Electronic Properties of Materials
Electron transport: phonons andelectron-phonon scattering
Reading:
I Kasap: 4.10
1
Phonons in 1D
I 1D chain of atoms of mass M connected by springs K
I Frequencies ω = 2√
KM
∣∣sin ka2
∣∣I Group velocity at k → 0 is vL = a
√KM , the sound velocity
0
2(K/M)1/2
-π/a 0 +π/a
Fre
quen
cy ω
Wavevector k
2
Phonons in 3D (eg. Au)
0
10
20
Γ X W L Γ K
Ener
gy [m
eV]
I Phonons have a band structure, just like electrons
I Key difference: linear near k = 0(Γ) rather than quadratic
I In 3D isotropic mateirals: two sound speeds: longitudinal (vL) andtransverse (vT )
I Two tranverse modes degenerate near k = 0 only (three polarizations)
I Compare energy scale against electrons: factor of 100 smaller
3
Phonon density of states
I Number of states per unit volume in ~k per unit volume = 1(2π)d
I Phonon dispersion relations ω = ωn(~k) (many bands)
I Therefore density of states g(ω) =∑n
∫d~k
(2π)dδ(ω − ωn(~k))
I Need a simple model to evaluate analytically (analogous to free electrons)
4
Debye model
I How many total phonon states in one band per unit cell? One!
I Debye model: keep linear dispersion with correct total number of states
I If ω = v|~k| for one band, density of states
g(ω) =
∫d~k
(2π)dδ(ω − v|~k|)
=
∫xdk
d−1dk
(2π)dδ(ω − vk)
=
∫xdw
d−1dw
(2πv)dδ(ω − w) (w = vk)
=xdω
d−1
(2πv)d
=4πω2
(2πv)3(in 3D)
5
Debye frequencyI For linear phonon dispersion in 3D
g(ω) =4πω2
(2πv)3
I If this is true for all ω, then total number of modes is∫
dωg(ω) = ∞I Debye model: cutoff model at ωD to keep correct number of states
I Number of states per unit volume (one band) = nion (# ions / volume)
I Therefore impose condition:
nion =
∫ ωD
0
dωg(ω) =
∫ ωD
0
dω4πω2
(2πv)3
=4πω3
D
3(2πv)3=
ω3D
6π2v3
⇒ ωD = v(6π2nion)1/3
I Corresponding cutoff in wave vector, kD = ωD/v = (6π2nion)1/3
I Note similarity to kF = (3π2n)1/3 for electrons
6
Phonon density of states: real metals
60
120
180
240
300
DO
S [10
29 e
V-1
m-3
]a) AlEq. 4 (this work)
Debye
60
120
180
240
300b) Ag
0
60
120
180
240
0 0.02 0.04 0.06
DO
S [10
29 e
V-1
m-3
]
ε [eV]
c) Au
0
60
120
180
240
0 0.02 0.04 0.06
ε [eV]
d) Cu
I Using two Debye models for velocities vL and vTI Note energy scale ED = kBTD = ~ωD
Phys. Rev. B 91, 075120 (2016)
7
Bose statistics
I At temperature T each phononic state of energy E has average occupation
nph(E) =1
exp EkBT− 1
I For massless bosons like photons and phonons, µ = 0 because number notconstrained
0
1
µ-2kBT µ µ+2kBTE
f(E)Bose
Classical
Fermi
8
Lattice heat capacity
I Density of phonon modes g(ω) = 4πω2
(2πv)3 Θ(ωD − ω) (Debye model)
I Phonon modes occupied by Bose function nph(~ω) at temperature T
I Internal energy per unit volume:
U =
∫ ∞0
dω(~ω)g(ω)nph(ω)
=
∫ ∞0
dω(~ω)4πω2
(2πv)3Θ(ωD − ω)
1
exp ~ωkBT− 1
=~
2π2v3
∫ ωD
0
ω3dω
exp ~ωkBT− 1
9
Lattice heat capacity: high T (T TD)
I At high temperatures ~ω kBT ⇒ exp ~ωkBT− 1 ≈ ~ω
kBT
I Correspondingly, internal energy is approximately
U ≈ ~2π2v3
∫ ωD
0
ω3dω~ωkBT
=1
2π2v3
∫ ωD
0
ω2dωkBT
=1
2π2v3ω3D
3kBT
=k3D6π2
kBT = nionkBT
I Accounting for the three polarizations, U = 3nionkBT and CV = 3nionkBI This is (of course) the equipartition result: high T → classical limit
10
Lattice heat capacity: low T (T TD)I At low temperatures,
U ≈ ~2π2v3
∫ ωD
0
ω3dω
exp ~ωkBT− 1
=~
2π2v3
(kBT
~
)4 ∫ ~ωDkBT
0
x3dx
ex − 1x ≡ ~ω
kBT
≈ k4BT4
2π2~3v3
∫ ∞0
x3dx
ex − 1
~ωDkBT
=TDT 1
=k4BT
4
2π2~3v3· π
4
15=π2k4BT
4
30~3v3
I Accounting for one longitudinal and two transverse velocities:
U = T 4π2k4B
30~3
(1
v3L+
1
v3T1
+1
v3T2
)∝ T 4
I And corresponding heat capacity
CV ≡dU
dT= T 3 2π2k4B
15~3
(1
v3L+
1
v3T1
+1
v3T2
)∝ T 3
11
Lattice heat capacity: real metals
Debye model does a good job! Proportional to T 3 for T TD,constant 3nionkB for T TD (classical equipartition result)
10
20
30
40
Cl [
10
5 J
/m3K
] a) Al
Eq. 5 (this work)Debye
10
20
30
40b) Ag
0
10
20
30
0 0.5 1 1.5 2
Cl [
10
5 J
/m3K
]
Tl [103 K]
c) Au
0
10
20
30
0 0.5 1 1.5 2
Tl [103 K]
d) Cu
Phys. Rev. B 91, 075120 (2016)
12
Heat capacity: electrons vs lattice
Mostly lattice! Except at very low T (< 10 K) when T wins over T 3
102
103
104
105
106
C [
J/m
3K
]
a) Al
TotalLattice
Electronic 102
103
104
105
106 b) Ag
102
103
104
105
106
100
101
102
103
C [
J/m
3K
]
T [K]
c) Au
102
103
104
105
106
100
101
102
103
T [K]
d) Cu
13
Electronic thermal conductivity setupI Energy current due to one electron E~v
(contrast with −e~v for charge current)
I At energy E, energy flux g(E)f(E)(Ev(E)) in random directions,averages to zero
I Non-uniform temperature ⇒ average energy flow in one direction
I Assume constant temperature gradient T (x) = T0 + xdTdx
I What is the net flux of energy across x = 0?
qx =
∫dx
∫dEg(E)f(E, T (x))(Ev(E))
·
∫ 1
02πd cos θ
4π · e−(−x/ cos θ)/λ
λ , x < 0
−∫ 0
−12πd cos θ
4π · e−(−x/ cos θ)/λ
λ , x > 0
14
Electronic thermal conductivity derivation
qx =
∫dx
∫dEg(E)f(E, T (x))(Ev(E))
∫ 1
0d cos θ
2 · ex/(λ cos θ)
λ , x < 0
−∫ 0
−1d cos θ
2 · ex/(λ cos θ)
λ , x > 0
=
∫dx
∫dEg(E)
(f(E, T0)︸ ︷︷ ︸∫→0 (equilibrium)
+∂f(E, T
∂Tx
dT
dx
)(Ev(E))
·
∫ 1
0d cos θ
2 · ex/(λ cos θ)
λ , x < 0∫ −10
d cos θ2 · e
x/(λ cos θ)
λ , x > 0
=dT
dx
∫dEg(E)v(E)E
∂f(E, T )
∂T
∫ 1
0d cos θ
2
∫ 0
−∞ dxx ex/(λ cos θ)
λ
−∫ −10
d cos θ2
∫∞0
dxx ex/(λ cos θ)
λ
=dT
dx
∫dEg(E)v(E)E
∂f(E, T )
∂T
−∫ 1
0d cos θ
2 λ cos2 θ
−∫ −10
d cos θ2 λ cos2 θ
= −dT
dx
∫dEg(E)v(E)E
∂f
∂Tλ/3
κ =τ
3
∫dEg(E)v2(E)E
∂f(E, T )
∂T(λ = vτ)
15
Electronic thermal conductivity: result
Substitute Fermi function in thermal conductivity expression:
κ =τ
3
∫dEg(E)v2(E)E
∂f(E, T )
∂T
≈ v2F τ
3
∫dEg(E)E
∂f(E, T )
∂T
(∂f
∂Tsharply peaked
)=v2F τ
3CV =
v2F τ
3· g(EF )
π2k2BT
3
=v2F g(EF )τ
3· π
2k2BT
3
Note similarity to electrical conductivity σ =v2F g(EF )τ
3 · e2.Therefore expect Lorenz number
L ≡ κ
σT=π2k2B3e2
≈ 2.44× 10−8(V/K)2
to be same temperature-independent constant across metals(not just free-electron-like ones)
16
Wiedemann-Franz law
L ≡ κ
σT=π2k2B3e2
≈ 2.44× 10−8(V/K)2
Metal L at T = 100 K (V/K)2 L at T = 273 K (V/K)2
Copper 1.9× 10−8 2.3× 10−8
Gold 2.0× 10−8 2.4× 10−8
Aluminum 1.5× 10−8 2.2× 10−8
Zinc 1.8× 10−8 2.3× 10−8
Lead 2.0× 10−8 2.5× 10−8
17
Lattice thermal conductivityI Phonons cannot transport charge, but they can transport heat.
I Our derivation for electrons didn’t assume electrons till this point:
κe =τ
3
∫dEg(E)v2(E)E
∂f(E, T )
∂T
I Lattice contribution: phonon DOS and Bose occupations instead
κL =τph3
∫dωg(ω)v2(ω)~ω
∂nph(ω, T )
∂T
I Which is larger?
I For semiconductors / insulators: few electrons (and holes)⇒ κL dominates
I For metals:I n ≈ nion
I vF ∼ 106 m/s vL, vT ∼ 103 − 104 m/s (electrons win)I τph ∼ ps, while τe ∼ 10 fs (phonons win)
I Net result: κe dominates at room temperature, κL important at high T
I Small κL ⇒ Lorenz number (based only on κe) close to ideal
18
Electron-phonon scattering rateI All transport coefficients depend on scattering time τ
I For electrons: τ determined by electron-phonon scattering
I Crude argument during Drude discussion:I Electrons only scatter against displaced ions (now we know why: band
theory)I Cross-section ∝ mean-squared displacement ∝ kBT (equipartition)I Therefore τ−1 ∝ T , and ρ = m
ne2τ∝ T
I Why is this not true at low T?
I Equipartion no longer valid for T TD
0 100 200 300
Temperature [K]
0
1
2
3
4
Resi
stiv
ity [
10
-8 m
]
Pure copperCold worked
+ 1% Ni
+2% Ni
19
Electron-phonon scattering rate estimationI Process: electron in one state absorbs (or emits) a phonon to go into
another state
I Phonon absorption rate of electron in state i (Fermi’s Golden rule):
τ−1 =2π
~∑j
|〈j|H ′|i〉|2δ(Ej − Ei − ~ω)
(using j for final state to prevent mix up with Fermi energy EF )
I Ingredient 1: sum over states∑j
I Count up electronic states using∫
dEjg(Ej)
I State must be unoccupied (Fermi statistics): (1− f(Ej))(was not relevant before because only one electron)
I Also need to sum over phonons available to absorb:∫
dωg(ω)nph(ω)
I Net result:∑j
→∫
dEjg(Ej)(1− f(Ej))
∫dωg(ω)nph(ω)
20
Electron-phonon scattering: selection ruleI So far assumed that final states counted as:∑
j
→∫
dEjg(Ej)(1− f(Ej))
∫dωg(ω)nph(ω)
I But many pairs of final states will not satisfy ~~ki + ~~kph = ~~kjI Instead of all pairs, only count those with correct momentumI RHS on sphere of radius kj ≈ ki (phonon energies negligible)
I LHS on sphere of radius kph centered on ~kiI Total pairs of ~kj ,~kph: (4πk2j ) · (4πk2ph)I Momentum conserving set: 2πkphI Fraction conserving momentum:
2πkph(4πk2j ) · (4πk2ph)
=1
8πk2jkph≈ vph
8πk2i ω
I Net final state counting:∑j
→∫
dEjg(Ej)(1− f(Ej))
∫dωg(ω)nph(ω)
vph8πk2i ω
21
Electron-phonon scattering: momentum relaxation
I Does electron-phonon scattering randomize momentum?(We assumed this about τ in Drude theory and its quantum extension.)
I Change in angle θ between ~ki and ~kj given by
cos θ =k2i + k2j − k2ph
2kikj≈ 1−
k2ph2k2i
I Extent of randomization proportional to:
1− cos θ ≈ ω2
2k2i v2ph
I To calculate momentum-relaxtion time instead ofmean-free time, include this (1− cos θ) factor:
∑j
→∫
dEjg(Ej)(1− f(Ej))
∫dωg(ω)nph(ω)
vph8πk2i ω
· ω2
2k2i v2ph
22
Electron-phonon scattering: matrix elementI Matrix element:
〈j|H ′|i〉 =
∫d~rψ∗j (~r)∆Vph(~r)ψi(~r)
where ∆V (~r) is change in electron potential due to phonon
I For estimate, make similar to argument made in classical case
I Instead of equipartition, apply to single phonon mode
I Mean-squared atom displacement ∝ ~ωI ∆V (~r) ∝ displacement of atoms (not squared)
I Therefore assume
〈j|H ′|i〉 ∝√~ω ≈
√ce-ph~ω(say)
(constant of proportionality ce-ph has dimensions of energy)
23
Electron-phonon scattering: Fermi golden ruleI Collecting ingredients together:
τ−1 ≈ 2π
~
∫dEjg(Ej)(1− f(Ej))
∫dωg(ω)nph(ω)
· vph8πk2i ω
· ω2
2k2i v2ph
·(√
ce-ph~ω)2δ(Ej − Ei − ~ω)
I Neglect phonon energies ~ω Ej , Ei (two orders smaller)
τ−1 ≈ 2π
~
∫dEjg(Ej)(1− f(Ej))
∫dωg(ω)nph(ω)
· ce-ph~ω2
16πk4i vphδ(Ej − Ei)
=2π
~g(Ei)(1− f(Ei))
∫dω g(ω)︸︷︷︸∝ω2
nph(ω)ce-ph~
16πk4i vphω2
∝ g(Ei)
∫ ωD
0
dωω4
exp ~ωkBT− 1
24
Electron-phonon scattering: temperature dependenceI For high temperatures T TD, exp ~ω
kBT− 1 ≈ ~ω
kBT:
ρ ∝ τ−1 ∝ g(EF )
∫ ωD
0
dωω4
~ωkBT
∝ g(EF )T
I For low temperature T TD,
ρ ∝ τ−1 ∝ g(EF )
∫ ωD
0
dωω4
exp ~ωkBT− 1
≈ g(EF )
(kBT
~
)5 ∫ ∞0
dxx4
ex − 1
∝ g(EF )T 5
25
Electron-phonon scattering: conclusions
I Temperature dependence:
ρ ∝ τ−1 ∝ g(EF )
T, T TD
T 5, T TD
explains low temperature deviations
I Reduction from classical result because few phonon quantaoccupied at low T (similar to blackbody spectrum reason)
I Transition metals ⇒ high d-band DOS at EF ⇒ high resistivity
0 100 200 300
Temperature [K]
0
1
2
3
4
Resi
stiv
ity [
10
-8 m
]
Pure copperCold worked
+ 1% Ni
+2% Ni
26
DFT + Fermi Golden ruleQuantitative prediction of resistivities
Metal DFT τ [fs] DFT ρ [Ωm] Expt ρ [Ωm]
Al 11.5 2.79× 10−8 2.71× 10−8
Cu 35.6 1.58× 10−8 1.71× 10−8
Ag 36.4 1.58× 10−8 1.62× 10−8
Au 26.3 2.23× 10−8 2.26× 10−8
Far from isotropic even for nominally free electron metals
ACS Nano 10, 957 (2016)
27